Question 171
Let $ABC$ be a right-angled triangle with $\hat{A}=90^{\circ}$ and $\hat{B}=30^{\circ}$. The perpendicular at the midpoint $M$ of $BC$ meets the bisector $BK$ of the angle $\hat{B}$ at the point $E$. The perpendicular bisector of $EK$ meets $AB$ at $D$. Prove that $KD$ is perpendicular to $DE$.